Question

Show that $\int x\sin x\cos x=f\left(x\right)$, taking const. of integration as zero. Find $f\left(\pi /4\right)$

Answer 1

$I=\int x\sin x\cos xdx$

$=\frac{1}{2}\int x\sin 2xdx$

Applying integration by parts
$=\frac{1}{2}[x(-\frac{\cos 2x}{2})-\int 1.(-\frac{\cos 2x}{2})dx]$

$=-\frac{1}{4}x\cos 2x+\frac{1}{4}.\frac{\sin 2x}{2}+C$

$=-\frac{1}{4}x\cos 2x+\frac{1}{8}\sin 2x$ (Given $C=0$)

Hence, $f\left(x\right)=-\frac{1}{4}x\cos 2x+\frac{1}{8}\sin 2x$

Hence $f\left(\frac{\pi }{4}\right)=-\frac{1}{4}\frac{\pi }{4}\cos 2\frac{\pi }{4}+\frac{1}{8}\sin 2\frac{\pi }{4}=\frac{1}{8}$